Complex Analysis3U Complex Methods/3H Methods in Complex Analysis \html-tag:br{\html-attr:className{"linebreak"}} Lecture Notes

Christian Voigt

This course is mainly concerned with the differentiation and integration of complex functions (complex-valued functions defined on subsets of the complex plane). Differentiation is in many ways the same as for real functions, but there are some interesting differences. In particular, a complex function differentiable on an open disc or on the entire plane can always be differentiated infinitely often and be expressed as the sum of a Taylor series. Integration has a much richer theory than for real functions, because there are many paths in a plane from one point to another.

These lecture notes are based on material prepared by Richard Steiner and Christian Korff.

PART I. DIFFERENTIATION

Limits

Sequences of complex numbers and their limits are defined in basically the same way as for real numbers. By definition, a sequence of complex numbers is a function , usually written where .

We say that converges to if for every there exists , such that for any with , we have . Equivalently,

In this case we shall write .

A sequence of complex numbers converges to if and only if the real sequences and are convergent with limits and , respectively.

A particulary important role in complex analysis are played by infinite series. By definition, a series of complex numbers is an expression of the form

for complex numbers . The series is called convergent if the sequence of partial sums converges. In this case the limit is called the sum of the series. Many properties of real series, including criteria for checking convergence, carry over to the complex case.

Let be such that . Then the geometric series

converges, and its sum is .

We shall also be interested in limits of functions.

Let be a subset of , let be a function from to , and let . We say that converges to as tends to if
(1) there exists such that


(2) for all there exists such that if then


In this case we write .

Note that need not be contained in . Limits of functions can be described in terms of sequences, in analogy to the real case.

Let be a subset of , let be a function from to , and let . Assume that there exists such that .
Then converges to as tends to if and only if for every sequence in with and for all we have .

Introduction to differentiation

CMS1

Differentiation of complex functions is defined in essentially the same way as over the real numbers.

CMD1.1 Let be a subset of , let be a function, and let be a point of . One says that is differentiable at , with derivative , if

  1. there exists such that

  2. for all there exists such that if then

In other words, is differentiable at , with derivative , if is defined at all points sufficiently close to and if the difference quotient

converges to when tends to . As usual, if the derivative of at exists, then it is denoted . If is differentiable at all points of we also write

for the derivative, viewed as a function on with values in .

Many results regarding differentiability for complex functions are the same as for real functions. For instance, if is a constant then it is easy to see that

The following rules are verified in the same way as in real analysis.

If and are differentiable functions then

By using these results, one can differentiate polynomials and rational functions, just as one would expect.

If is a natural number, then

This follows inductively by writing and using the product rule.

The chain rule holds as well. That is, if is differentiable at and is differentiable at , then is differentiable at and

CME1.3 Show that the function is not differentiable anywhere on .Let be arbitrary. Let be a nearby point with . Then for some , and the difference quotient is

The difference quotient takes arbitrary values with modulus for arbitrarily small values of , so it does not tend to a limit as tends to . Therefore the function is not differentiable at .CME1.4 At which points is the function differentiable?Let be any point and let be a nearby point. Then the difference quotient is

In this expression tends to as tends to . If , then takes arbitrary values of modulus with arbitrarily close to , so the difference quotient does not tend to a limit. But if , then the difference quotient tends to . Therefore is differentiable only at .

This is a curiosity; we are really interested in functions which are differentiable on large regions, not just at isolated points.

CMD1.5 Let be a point in and let be a positive real number. Then the open disc with centre and radius is the set

CMD1.6 A subset of is open if for every point there is an open disc with centre entirely contained in .CME1.7 Every open disc is open, and is open.CMD1.8 A holomorphic function is a function such that is an open subset of and is differentiable at every point of .

This course is about holomorphic functions. Note that, for a complex function defined on an open set , the first condition for differentiability in Definition CMD1.1 is automatically satisfied at every point of .

Power series

CMS2

Complex series and power series behave just like real ones. By a complex power series we mean a series of the form

where and are complex numbers. Let us list some important facts about such power series.

power Let be a complex power series.

  1. There is a radius of convergence with such that the series converges for and diverges for .

  2. Suppose that and let be the open disc with centre and radius , with the convention that if . Let be the sum of the series. Then is holomorphic on . The derivative is given by

    and also has radius of convergence .

In many cases the radius of convergence can be found from the ratio test.

CMT2.2 If a power series

is such that converges to a finite or infinite limit as , then the radius of convergence is .

Here we use the convention and .

Assume now that is a complex power series with radius of convergence , and let be the open disc of radius centred at . By applying Theorem power repeatedly, one sees that the function can be differentiated infinitely often on and that

The original series is then equal to the Taylor series

There are also functions on such that , in other words, we can always find an antiderivative of . These functions are given by the series

where is an arbitrary constant, and one can show that these series also have radius of convergence .

CME2.1 From the theory of geometric series, there is a power series in with radius of convergence given by

such that . By differentiating and integrating term by term, we obtain power series and with radius of convergence such that

and such that

By modifying this example, if is any non-zero complex number, then has an antiderivative on the open disc with centre and radius .

Elementary functions

CMS3

The exponential function can be defined for all complex numbers as the sum of the power series

the radius of convergence being infinite by the ratio test. One can then define and for all complex numbers by

and

the radii of convergence again being infinite. These definitions agree with the usual definitions when is real, and as in the real case one has

We note also that Euler’s identity

holds for all complex numbers .

CME3a For any fixed the function is constant.For one has

using the product rule and the above formula for the derivative of . It follows that

for all , which yields the claim.

In particular, Lemma CME3a shows , so is non-zero for all , and

For all complex numbers and it now follows from the formula that

In particular for with we have

so that the complex exponential function can be expressed in terms of the real exponential and trigonometric functions. The complex sine and cosine functions can then also be expressed in terms of the real exponential and trigonometric functions, as can the other trigonometric functions like

Moreover we can easily recover various known properties of trigonometric functions.

CME3.0 Show that for all .We calculate

directly from the definition of and .

We also have the hyperbolic sine and cosine functions defined by

and other hyperbolic functions like . These hyperbolic functions are just slight variants of the usual trigonometric functions, in the sense that

for all .

CME3.0.5 Show that for all .This is similar to Example CME3.0. We calculate

directly from the definition of the hyperbolic sine and cosine functions.

From Example CME3.0.5 it follows in particular that the points for form the unit hyperbola

This is the reason why these functions are called hyperbolic. By comparison, note that the set of all for is the unit circle, which is why the functions and are called trigonometric.

Let us next discuss logarithms. For a given complex number we consider the equation

The solutions of this equation are called the complex logarithms of . If then there are no solutions, since we have seen above that is nonzero for every . For let

be a polar form of , and let

be the cartesian form of ; then

We write this in the form

Here is the ordinary real natural logarithm of the positive real number , and is the argument of .

Note that has infinitely many values: if is one value then the complete set of values is given by where is any integer. Correspondingly, has infinitely many values. The notation should therefore be used with great care.

CME3.1 Find the complex logarithms of for .We have

For with we have and

where is an arbitrary integer. In this case and

so

where is an arbitrary integer.

If we restrict attention to some fixed branch of the argument, like we can turn into a (well-defined) holomorphic function. The equation

can then be differentiatedto give

hence

for all . In this way we get antiderivatives for the function , at least locally.

We use logarithms to define generalised powers for with by the formula

These are again multivalued, so should be treated very cautiously. However, they do give holomorphic functions on small regions if one makes a continuous choice for the logarithm. In general, one gets an infinite sequence of possible values for , but if is a rational number such that and are coprime integers and , then one gets precisely distinct values. In particular, if is an integer then has a single value. These single values are as one would expect:

CME3.2 Calculate all values for .We have , and for any integer. Therefore we obtain

as solutions.

Using the above methods we can also solve equations involving functions which are built out of the exponential function, like trigonometric functions.

CME3.2 Find the complex numbers such that .We treat the equation

as a quadratic equation in . After rearrangement it becomes

The solutions are

so that

that is,

where is an arbitrary integer.

The Cauchy–Riemann equations

CMS4

By taking real and imaginary parts, a complex-valued function of a complex variable can be regarded as a complex-valued function of two real variables or as a pair of real-valued functions of two real variables. It turns out that the differentiability of the complex-valued function is essentially equivalent to two simultaneous equations between the partial derivatives of the real-valued functions.

CMT4.1 Let be a complex-valued function of a complex variable and let

so that

If is differentiable with respect to , then the real-valued functions and have (continuous) partial derivatives with respect to and and

Conversely, if and are real-valued functions having continuous partial derivatives with respect to and such that the Cauchy–Riemann equations

hold, then is differentiable with respect to .Suppose first that is differentiable with respect to . Then for any given value of we have

Restricting to real values shows that

Restricting to purely imaginary values shows that

Equating real and imaginary parts in the two expressions for therefore gives

as required.
Conversely, suppose that and have continuous partial derivatives satisfying the Cauchy–Riemann equations. There are then continuous real-valued functions and of such that

For any given value of we will show that

as ; this will show that is differentiable as a function of , with derivative .
Let with . Take in the form

and expand this to obtain

By the mean value theorem from real analysis we find such that

where and are between and , and and are between and . Note here that if we may take , and similarly if . The points depend on , and they tend to as . On the other hand we have

Therefore

Here because . Moreover we have as because is continuous, and similarly for the other terms in this expression. It follows that

as , as required.
This completes the proof.CME4.2 Let

Then and have continuous partial derivatives with respect to and , and

so there is a holomorphic function given by

with

This gives an alternative approach to the complex exponential function.CME4.3 Let and ; then , so the function

is not differentiable. This gives an alternative proof that the complex conjugate function is not holomorphic.

Harmonic functions

CMS5

A harmonic function is a real-valued function of two real variables and which satisfies the Laplace equation

Harmonic functions are important in certain branches of applied mathematics.

CMT5.1 If and satisfy the Cauchy–Riemann equations and have continuous second-order partial derivatives, then they are harmonic.Using the Cauchy–Riemann equations and the symmetry of continuous second-order partial derivatives, we see that

and similarly for .

Let be a pair of real-valued functions with continuous second-order partial derivatives. If and satisfy the Cauchy–Riemann equations, then is called a harmonic conjugate of . On reasonable domains, like open discs,every harmonic function has a harmonic conjugate, so that harmonic functions are essentially the same as the real parts of holomorphic functions.

One can find a harmonic conjugate by solving the two Cauchy–Riemann equations successively.

CME5.2 Show that the function is harmonic, and find a harmonic conjugate.We have

It follows that is harmonic, because

For to be a harmonic conjugate, we require

that is,

The general solution to the first of these equations is

where is an arbitrary function in . For to be a solution to the second equation one then requires to be differentiable and

which means that , where is constant. Thus the harmonic conjugates are given by

where is constant.
Note that when performing the first antidifferentiation in the above argument one needs to be an arbitrary function, not just an arbitrary constant.

This example shows that there is a holomorphic function given by

In fact, by the binomial theorem this can be written as .

Heuristically, one can often find a formula for from the formula for by replacing with and replacing with ; this is because

PART II. INTEGRATION

Introduction to integration

CMS6

An important part of real analysis concerns the integration of functions over closed intervals of the real line. Over the complex numbers we can integrate functions in a similar way, but we now have additional flexibility in choosing the domains we want to integrate over.

We will be interested in the contour integration of functions in the complex plane. Essentially, if is a function of with real and if is a path in the complex plane then we will get

which means that we are dealing with pairs of real line integrals. We will, however, give the definitions independently in a reasonably general form.

CMD6.1 Let be a subset of . A path in is a continuous map where is a closed interval in .

Let be a path. We can approximate the length of by sums of the form

where

If these sums tend to a finite limit as tends to zero, then one says that has finite length .

If is of finite length and if is a continuous function, then the contour integral

is obtained from the Riemann sums

where

is a partition of , and where

by taking the limit as .

We will not give the details of this construction in this course. In the sequel, the paths involved in contour integrals are supposed to be of finite length, even if this is not stated.

Note that the family of Riemann sums is determined by the contour of , that is the points in the image of , and by the order in which they occur; the precise parametrisation does not matter. Because of this, we tend to specify paths by stating their images and the direction of travel. For example, we may refer to an integral along a straight line segment from to or an anticlockwise circular arc; this really means the segment or arc with any convenient parametrisation in the correct direction. Note that the direction of travel must be specified, because of the following result.

CMP6.2 If the direction of travel along a path is reversed, then the integrals along the path are multiplied by .The Riemann sums for the reverse direction are obtained from the original Riemann sums by multiplying by .

There is a very important and very useful bound for contour integrals, as follows.

CMT6.3 If is of finite length and if for all points in the image of , then

For each Riemann sum one has

and the result follows by taking the limits as tends to zero.

This result can be used to estimate integrals. The ability to do this is an intended learning objective for Methods in complex analysis but not for Mathematics 3U, and this is the only real difference in the intended learning objectives between the two courses. But the result should be known in any case, because it is frequently used in proofs.

In practice, contour integrals are converted into integrals over intervals in the real line, essentially by change of variables.

CMD6.4 A path on is a piecewise path if continuously differentiable at all but finitely many points of the interval .CMP6.5 Let be a piecewise path. Then has finite length if and only if the absolute value of derivative has a finite integral over . If has finite length, then the length is given by

If has finite length and if is a continuous function on , then

We will not give a proof of this result. Let us instead take a look at some examples.

CME6.6 Let be the path given by

for some and . Compute the length of .We have , so the length of is

CME6.7 Let be the straight line from to . Compute

We can parametrise as

for . Then and , so

Informally, we may write and substitute and in to calculate the integral.

In general the straight line from to can be parametrised as

for . Other parametrisations are possible; in particular one does not have to use the interval .

CME6.8 Let be the path given by

and let be an integer. Compute

We have

In cases with we get

and we get

in the case .

In order to calculate contour integrals we can often use what amounts to the fundamental theorem of calculus.

CMP6.9 Suppose is apath in an open set from to and let be a continuous function on such that for some differentiable function on , then

This gives another way to look at Example CME6.8 in the case : since is a path from to in we have

CMD6.10 A closed path in is a path such that .CMP6.11 If is a closedpath in and if is a continuous function on such that for some differentiable function on , then

CME6.12 If is a closedpath in and if is a polynomial, then

CME6.13 Let be the circle with centre and radius traced anticlockwise. By integrating along , show that there is no function on with derivative .We can parametrise as for . We have

hence

Since is a closed path, it follows that cannot have an anti-derivative on .

Cauchy’s theorem

CMS7

By the fundamental theorem of calculus, the integral along a closed path of a function with an anti-derivative is equal to zero, see Proposition CMP6.11. Cauchy’s theorem says that, under certain circumstances, one gets the same result for functions with derivatives rather than anti-derivatives.

We begin with the special case in which the domain of the closed curve is the boundary of a rectangle.

CMT7.1 Let be a differentiable function on an open set in , let be a rectangle in with boundary , and let be a path of finite length such that for some continuous function . Then

The function may be very far from smooth. However, since is open and is of finite length, we have enough wriggle room to change without changing its values on the boundary such that it multiplies lengths by a bounded amount. We assume that this has been done. In particular, let is a rectangle contained in with boundary traced in the same direction as ; then is of finite length and we can make the definition

In this notation, we want to show that .
Bisect horizontally and vertically into four rectangles ; then

because the integrals along edges inside cancel out. We can therefore choose one of the small rectangles, say, such that

By repeating this process, we can construct a sequence of rectangles

each a quarter of its predecessor, such that


Now let be the lower left hand corner of . The sequence is increasing and bounded above, so it converges to a limit , and similarly converges to a limit . Let ; then for all .
We now use the fact that is differentiable. For let

and let ; then is continuous on and

for all . Since the integral of a polynomial along a closed curve is zero (see Example CME6.12) we have


Now the linear dimensions of are obtained from those of by multiplying by . Since multiplies lengths by a bounded amount, there are positive constants and such that the length of is at most and such that for in . We also have as , so there are positive constants such that for in and such that as . By Proposition CMT6.3,

Since and since as , it follows that . This completes the proof.

We apply this result to show that in certain circumstances the integrals of a function along different paths are equal.

CMD7.2 Let

be two closed paths with the same domain and the same codomain . Then and are homotopic (as closed paths in ) if there is a continuous function

such that

The function is called a homotopy from to .

In this definition, the homotopy is a continuous family of closed paths , one for each point in , starting with and finishing with .

CMT7.3 Let be a differentiable function on an open set in and let and be closed paths in of finite length which are homotopic as closed paths in . Then

By a wriggle room argument, we can find a homotopy whose restriction to the boundary of its domain is of finite length. The integral of along its boundary in the appropriate direction is then

since the integrals along the other two sides cancel out because the homotopy is a homotopy through closed paths. The result therefore follows from Theorem CMT7.1.

We use this to generalise Theorem CMT7.1 as follows. Let us say that a closed path in an open set is null-homotopic in if it is homotopic to a constant path as a closed path in . Then the following result is an easy consequence of Theorem CMT7.3 because an integral along a constant path is obviously equal to zero.

CMT7.4 Let be a differentiable function on an open set in and let be a closed path in which is null-homotopic in . Then

In particular we can apply this result to paths which do not cross themselves.

CMD7.5 A simple closed path on an interval is a closed path such that for .

Thus a simple closed path is a closed path which is as near to being injective as possible. Note that for a closed path we have by definition, which means that a closed path cannot be injective.

Let be a simple closed path in . Then the complement of the image of is the union of two disjoint non-empty connected open sets, one of them bounded, one of them unbounded; this is the famous Jordan curve theorem. The members of the bounded set are inside , and the members of the unbounded set are outside . If is an open set containing and all the points inside , then one can show is homotopic to a constant path in . As a special case of Theorem CMT7.4 we therefore get the following result.

CMT7.6 Let be a simple closed path of finite length in , let be an open set containing and all the points inside , and let be differentiable. Then

Cauchy’s integral formulae

CMS8

The theorems in the previous section are very powerful. We will now describe some of their consequences.

If is a simple closed path in then may be traced anticlockwise or clockwise. To be more precise, let be a set containing and all the points inside , and let be a point inside . One can show that is homotopic in to a small circle with centre , and one can show that the circles for the various points are all traced the same way, either anticlockwise or clockwise. We say that itself is traced anticlockwise or clockwise accordingly. We will mainly consider paths that are traced anticlockwise, because of the following theorem.

CMT8.1 Let be a simple closed path in traced anticlockwise, let be an open set containing and all the points inside , let be differentiable, and let be a point inside . Then

If we write

then we must show that .
Let be a circle with centre and with small radius traced anticlockwise. If is sufficiently small, then is homotopic to in . The function is differentiable on , so

by Theorem CMT7.3.
We now consider what happens if is replaced by in the integral . We can parametrise as with running from to , and we get

It follows that

Now the difference quotient converges to as tends to ; therefore, if is sufficiently small we get

for all in . The length of is , so for such it follows from Proposition CMT6.3 that

Since can be arbitrarily small we conclude that as required.

In this theorem, the integrand is infinitely often differentiable with respect to for each fixed in the path . One can deduce that is infinitely often differentiable inside , with the obvious derivatives. This gives the following generalisation of Theorem CMT8.1.

CMT8.2 Let be a simple closed path in traced anticlockwise, let be an open set containing and all the points inside , and let be differentiable. Then can be differentiated infinitely often at any point inside , and the derivatives for are given by

For every point in an open set there is a disc in with centre . By considering a circle in this disc with centre , one gets the following result.

CMT8.3 Let be an open set in and let be a differentiable function. Then can be differentiated infinitely often.

This is a very significant difference between complex analysis and real analysis. In fact, in real analysis there are differentiable functions whose derivative is not even continuous, let alone differentiable. A basic example of this phenomenon is the function given by

This function is differentiable in the sense of real analysis with

This can be calculated using the product rule and chain rule as usual if , and directly from the definition at .Note that is discontinuous at .

Applications

CME9

We consider holomorphic functions on open discs and on the entire complex plane . These functions are not merely infinitely differentiable, as shown in Theorem CMT8.3; they are equal to the sums of their Taylor series.

In general, if is an open set then a function is called analytic if it can be locally expressed as a sum of convergent power series. We have already seen in Theorem power that analytic functions are holomorphic. Now we shall establish the converse.

CMT9.1 Let be a holomorphic function on an open set , let and let be an open disc with centre . Then is analytic and

for all in .Let

we must show that for . Let be a circlein with centre traced anticlockwise such that is inside . From Cauchy’s integral formula for derivatives, see Theorem CMT8.2, one can deduce that

here the interchange of integration and summation is justified because one can check that the series

is convergent.
Consider now the expression

Since is inside , we have

for each on , so the integrand is an absolutely convergent geometric series with sum given by

Therefore

and Cauchy’s integral formula in Theorem CMT8.1 implies that as required.

Theorem power and Theorem CMT9.1 show that a function defined on an open set is analytic if and only if it is holomorphic. This is why these two notions are often used interchangeably.

For holomorphic functions on the entire complex plane there is a further key result due to Liouville, given as follows.

CMT9.2 Every non-constant holomorphic function on the entire complex plane is unbounded.Let be a bounded holomorphic function, say

for all . We will show that is constant by showing that for all .
Indeed, let be a circle with centre and radius traced anticlockwise. By Cauchy’s integral formula for derivatives, see Theorem CMT8.2, we have

If is a point on then

Since the length of is , it follows from Proposition CMT6.3 that

This is true for all positive numbers , so as required.CME9.3 Let be an holomorphic function on such that for all . Show that is constant.Let ; then is a holomorphic function on such that for all . By Liouville’s theorem, is constant. Since is continuous, is constant.

Laurent series

CMS10

Let be a point in , let be an open disc with centre or let , let be a holomorphic function on , and let be a circle in with centre traced anticlockwise. By Cauchy’s integral formula for derivatives, Theorem CMT8.2, we have

for all . The Taylor series

which is valid for all in by Theorem CMT9.1, therefore takes the form

with

In this form, there is a simple generalisation to functions which are holomorphic only on : we use the terms for all integers , both positive and negative. The resulting series is like a power series, except that it involves both positive and negative powers. Series of this kind are called Laurent series; thus a Laurent series is a series of the form

One can write a Laurent series as the sum of a power series in and a power series in ,

by definition, the Laurent series is convergent if and only if these two power series are both convergent.

CMT10.1 Let be a point in , let be an open disc with centre or let , let be a holomorphic function on , let be a circle in with centre traced anticlockwise, and let be a point in . Then

where

for all .Let

we must show that


To evaluate , let be a circle in with centre traced anticlockwise such that is inside . Then is homotopic to in ; hence, by Theorem CMT7.3,

As in the proof of Theorem CMT9.1, it follows that


To evaluate , let be a circle in with centre traced anticlockwise such that is outside . We now get

The series in the integrand is a convergent geometric series because is outside , so that

for all on . The sum is now given by

so that


Recall now from Proposition CMP6.2 that changing the sign of a contour integral is equivalent to reversing the direction of travel. Let be a closed path in consisting of the circle traced anticlockwise, a path from a point on to a point on , the circle traced clockwise, and the path traced in reverse; then we see that

We now shrink this path slightly to get an anticlockwise simple closed path homotopic to in such that is inside . By Theorem CMT7.3,

hence, by Cauchy’s integral formula, see Theorem CMT8.1, we get as required.

Singularities

CMS11

Let be a holomorphic function on , where is an open set containing . In this case one says that has an isolated singularity at . By Theorem CMT10.1 there is a Laurent series expansion

for all sufficiently close to .

CMD11.1 With and as above there are the following three mutually exclusive possibilities.

  1. Suppose that for all . Then the series is a power series in the ordinary sense, with a non-zero radius of convergence. In particular it has the sum at . One can therefore extend to a holomorphic function on by putting . One says that has a removable singularity at .

  2. Suppose that is non-zero for for some negative values of , but only for finitely many, so that the Laurent series has the form

    with and . One says that has a pole of order at . In this case,

    for some function which is holomorphic on and non-zero at .

  3. Suppose that is non-zero for infinitely many negative values of . Then has an isolated essential singularity at .

Let us consider examples for all three types of isolated singularities as introduced in Definition CMD11.1.

CME11.4 The function

has a removable singularity at and can be made holomorphic at by giving it the value at that point. We have as , so is bounded on some neighbourhood of .
The function

has a pole of order (a simple pole) at , and a pole of order (a double pole) at . Obviously when or .
The function

has an isolated essential singularity at . If through positive real values, then ; if through negative real values then . For tending to through arbitrary complex numbers, is not bounded but does not tend to infinity either.

Singularities can always be classified by the behaviour of the absolute value as in these examples.

CMT11.5 Let be an open set and let be a holomorphic function on , so that has an isolated singularity at .

  1. The singularity is removable if and only if is bounded on some neighbourhood of .

  2. The singularity is a pole if and only if as .

  3. The singularity is an isolated essential singularity if and only if is unbounded on any neighbourhood of and does not tend to infinity as tends to .

(1) If the singularity is removable, then obviously is bounded on some neighbourhood of .
Conversely, suppose that is bounded on some neighbourhood, say for all sufficiently close to . Let the Laurent series of be

so that

for any anticlockwise circle with centre and with sufficiently small radius . If is in , then

and the length of is , so it follows from Proposition CMT6.3 that

If , then as , so . Therefore the singularity is removable.
(2) If the singularity is a pole, then we obviously have as . Indeed, if is a pole of order then we can write

and since , the behaviour of near is determined by the factor .
Conversely, suppose that as . Then there is an open disc with centre such that for . It follows that is defined on and is bounded on that set, so has a removable singularity at by part (1). Since as , and since for , looking at the Taylor series of at shows that there exists such that

with . Therefore with holomorphic and non-zero near , hence with holomorphic and non-zero near . Since we conclude that has a pole of at .
(3) This follows from parts (1) and (2).

The calculus of residues

CMS12

Assume is a holomorphic function on an open set . In the Laurent series expansion

see Theorem CMT10.1, the coefficient has the particularly simple expression

where is a suitable circle. This coefficient has a special name, and is of special importance for calculating integrals.

CMD12.1 If a holomorphic function has an isolated singularity at , then the residue of at , denoted , is the coefficient of in the Laurent series of at .CMT12.2 Let be an open set in , let be holomorphic, and let be an anticlockwise simple closed path in such that there are finitely many points inside not belonging to . Then

Let be disjoint open discs inside with their centres at . One can show that is homotopic in to a path consisting of arcs making up the boundaries of these discs traced anticlockwise and of connecting paths traced twice in opposite directions. The integrals along the connecting paths cancel out by Proposition CMP6.2. It therefore follows from Theorem CMT7.3 that

This yields the claim.

Theorem CMT12.2 generalises Cauchy’s Theorem CMT7.6, which is the case with no singularities, and Cauchy’s integral formulae, see Theorems CMT8.1 and CMT8.2, which are cases with one singularity.

In order to apply the residue theorem, it is necessary to calculate residues. This can sometimes be done by direct computations of Laurent series. For poles there is a useful general result as follows.

CMP12.3 If has a pole of order at most at then

In particular, if has a simple pole at then

Let , so that is holomorphic at , and let the Taylor series of at be . The Laurent series of is then given by

so that that . Therefore

as claimed.

Proposition CMP12.3 can sometimes be combined with l’Hôpital’s rule.

CMP12.4 If and are holomorphic function on an open set and for some then

Sometimes it is possible to use partial fractions to calculate residues.

CME12.5 Find the residues of the following functions at their singularities:

The singularities of are simple poles at . For each singularity , by Proposition CMP12.3,

By l’Hôpital’s rule,


The singularities of are a simple pole at and a double pole at . By Proposition CMP12.3,

and


The only singularity of is at . The Laurent series at is

so .
The singularities of are at and . The Laurent series can be found by adding the Laurent series for the individual partial fractions. The terms involving negative powers in the Laurent series are

so .CME12.6 Evaluate

where

and is the boundary of the square with vertices at traced anticlockwise.The singularities of inside are at and , so

using the results from Example CME12.5.CMR12.7 There is a generalisation of the residue theorem to arbitrary closed paths with essentially the same proof. If is any closed path in and if is a point not in the image of then there is a winding number

which is an integer and is the number of times that goes round in the anticlockwise direction. If is open in , if is holomorphic on , if is a closed path in , and if there are finitely many points with non-zero winding number not belonging to , then

Applications to integrals over real intervals

CMS13

The calculus of residues can be used to evaluate certain integrals of real functions over real intervals. One example is the famous Gaussian integral

but we will not explain how to derive this formula using residues here.

We will consider integrals of two other types. Firstly, integrals of the form

such that is periodic with period can sometimes be evaluated by the substitution .

CME13.1 Compute the integral

Make the change of variable , so that

Let be the circle with centre and radius traced anticlockwise; then

The integrand is holomorphic on and has one singularity inside , at . The residue at is

see Proposition CMP12.3. Therefore

by the residue theorem.

Let be a function such that is bounded for large values of . The integral

is convergent by comparison with

in fact

This limit can sometimes be computed by integrating round the half-discs

the boundedness condition on ensures that as , and from this it follows that the integrals along the semicircular parts of the boundaries of the half-discs tend to zero.

CME13.2 Compute the integral

Let

For , let

where is the boundary of the upper half of the disc with centre and radius . There are no singularities of on , and the singularities inside are at and . By Proposition CMP12.3,

therefore we get


The semicircular part of has length ; hence by Proposition CMT6.3 the absolute value along the semicircular part of is at most

This clearly tends to zero as tends to infinity, because is approximately for when is large. The integral along the straight line part of is

which tends to as tends to infinity, because is bounded for large values of . Therefore we get

for the value of the integral.

In examples of this kind, students in Methods in complex analysis must prove that the integral along the semicircle tends to zero as the radius tends to infinity; students in Mathematics 3U may assume this without proof.

PART III. CONFORMAL TRANSFORMATIONS

Introduction to conformal transformations

CMS14

Let be a bijective function between open sets in . We say that is conformal if preserves oriented angles between (the tangents to) intersecting curves. It turns out that this condition is equivalent to being holomorphic with everywhere non-zero derivative.

A function from to is an orientation-preserving isometry if and only if it is a composite of rotations and translations. Analogously, one can show that a function from to preserves oriented angles between straight lines if and only if it is a composite of rotations, translations and scalings, where a scaling is a function of the form

for some positive real constant . Given arbitrary points and , one can deduce that the functions from to preserving oriented angles between straight lines and taking to are the functions of the form

with , and with . In other words, they are the functions

with and with .

Conformal functions are functions of approximately this form in the neighbourhood of each given point; in other words, they are the holomorphic functions with everywhere non-zero derivatives.

One can show that a holomorphic function with non-zero derivative is locally a bijection. That is, if is holomorphic and for some then there exists an open neighborhood of and an open set such that is a bijection. We give the following definition.

CMD14.1 Let and be open subsets of . A conformal transformation from to is a bijective holomorphic function from to . One says that and are conformally equivalent if there is a conformal transformation from to .

One can show that an injective holomorphic function has everywhere non-zero derivatives. Since injective holomorphic functions have non-zero derivatives we have the following result.

CMP14.2 If is a conformal transformation, then for all .

Recall that an equivalence relation on a set is a binary relation which is reflexive, symmetric, and transitive. If we write the relation in the form for then this means

Let us verify that conformal equivalence is indeed an equivalence relation in a natural way.

CMP14.3 Conformal equivalence is an equivalence relation on the set of all open subsets of .Let be an open subset of . Then the identity is a conformal transformation from to , so is conformally equivalent to . This shows reflexivity.
To check symmetry suppose that is conformally equivalent to , so that there is a bijective holomorphic function . The inverse is a bijection such that

for all . Differentiating this equation implicitly gives

so that is differentiable with

this argument is valid because by Proposition CMP14.2. Therefore is conformally equivalent to .
Finally, to verify transitivity suppose that is conformally equivalent to and that is conformally equivalent to , so that there are conformal transformations

It follows from the chain rule that is a conformal transformation from to ; therefore is conformally equivalent to .
We conclude that conformal equivalence therefore has the three properties required for an equivalence relation.

If is a conformal transformation, then the holomorphic functions on are the functions such that is holomorphic on , hence the harmonic functions on are the functions such that is harmonic on , see Section CMS5. One can therefore, for instance, analyse fluid flows round a subset of in terms of flows round the possibly simpler image in .

CME14.4 Suppose that

Then the exterior of a circle passing through and with radius greater than in the -plane corresponds to the exterior of a set in the -plane resembling the outline of an aeroplane wing and called a Joukowski aerofoil. It has a cusp at the left and a vertical tangent at the right. The top is convex. The bottom is concave at the left and convex at the right. This idea is used in the design of aeroplanes.CME14.5 Find conformal transformations between the sets

There is a conformal transformation from to given by

with inverse

Note that is well-defined since

where .
There is a conformal transformation from to given by

corresponding to a choice of .
There is a conformal transformation from to given by .CME14.6 Find a conformal transformation from to , where

Note that if and only if . The exponential function converts bounds on imaginary parts into bounds on arguments, because is the argument of . The function converts bounds on real parts into bound on imaginary parts. Thus

and it follows that the transformation maps bijectively onto .CMT14.7 Let be an open set in . Then is conformally equivalent to an open disc if and only if and and all closed paths in are homotopic in .

A space in which all closed paths are homotopic is said to be simply connected.

Suppose that is conformally equivalent to an open disc.
By Liouville’s theorem, see Theorem CMT9.2, any holomorphic function on must be constant or unbounded, so its image cannot be an open disc; therefore .
It is obvious that .
It is clear that any two closed paths in a disc are homotopic; therefore the same must be true of .

The proof of sufficiency is much more difficult, and will not be covered in this course.

Möbius transformations

CMS15

A Möbius transformation is a function of the form

where are complex constants such that . The condition ensures that this function is a bijection from its domain to its image; the inverse is given by

If we get a bijection from to ; if we get a bijection from to ; in any case we get a conformal transformation. It is convenient to regard it as a bijection from to by making the definitions

(meaning in the case ). With this convention, Möbius transformations can be composed: if

then

where

CMP15.1 Under a Möbius transformation, the image of a set which is a circle or a straight line is again a set which is a circle or a straight line (with the convention that belongs to every straight line).The original set consist of the points satisfying a condition of the form

where and are distinct complex numbers and where is a positive real number. More precisely, for a straight line one needs , and for a circle one needs . Making the substitution gives a condition of the same form for . Indeed, we calculate

and multiplying both sides with gives a relation of the required form for .CMP15.2 Let be the open disc with centre and radius , and let be a conformal transformation such that . Then for some constant with .Consider the function , which is holomorphic on . We have as , so the singularity at is removable (see Theorem CMT11.5). Thus there is a holomorphic function on such that

for . We want to show that is a constant with absolute value . Let us first prove that for all .
For one can show that the maximum value of with is attained at some point with , see Question 5 on Exercise Sheet 5.For it follows that

using in the last inequality. Letting shows that for . Applying the same argument to shows that for all , hence for all as claimed.
We can now show that for some with . Decompose into its real and imaginary parts. Then from we get which implies

Inserting the Cauchy–Riemann equations, see Theorem CMT4.1, into the second equality gives

Multiplying this and the first equality by and , respectively, and adding up the results shows

It follows that

and similarly one shows that the partial derivatives of with respect to vanish. This means that and are constant, which finishes the proof.CMT15.3 Let be the open disc with centre and radius , and let be a conformal transformation from to . Then

for some constants and with and .Let , and let

Then is a conformal transformation from to such that . It follows that is a conformal transformation from to sending to ; therefore

for some constant with . Putting gives the result.CME15.4 Let

Find the image of under the transformation

We can use the idea of the proof of Proposition CMP15.1. Let

then

Let with ; then

The image of therefore consists of the points outside the circle with centre and radius .
To check this, note that

so that the boundary circle of is sent to the circle through and . This is the circle with centre and radius . It follows that the image of is either the region inside the image circle or the region outside the image circle. Since , the image is in fact the outside.CME15.5 Let be a Möbius transformation such that

Show that

for some real constants .We know that

for some complex constants such that .
Suppose that . Then and , so is a linear polynomial, say

with . We have and . Since and are real, the same is true for and .
Suppose that . Then we can write in the form

with . We have ; therefore . We have , so that ; therefore . Finally, we have , so that

therefore .